Question from textbook for practice. There is no answer, your input is needed.

CD is a firm that excavates roadside ditches for laying drainpipe with
a production function:
          2
Q=10L-0.1L 

(The above equation reads Q equals 10L minus 0.1L squared).

Where L denotes labour hours and Q the length of the ditch in meters.
The firm hires labour at a wage rate of $12 per hour.

1.	CD has received an offer to excavate 250 meters for a price of
$500. Should it accept the offer?
2.	Suppose CD receives a price of $2 per meter, what is CD?s maximizing profit?

Hi!!


1. CD has received an offer to excavate 250 meters for a price of
$500. Should it accept the offer?

Here you need to find the cost of the requested excavation, and see if
revenues are greater than costs to accept or reject the offer.

Revenue = $500 (this figure comes from the offer).
Cost = Labour hours needed to do the job * wage rate

To find the "Labour hours needed to do the job" use the Total
production function Q(L) to isolate L:
Q(L) = 10*L - 0.1*L^2 = 250  (this is the amount of meters that must be excavated)
Now we have that:
0 = 0.1*L^2 - 10*L + 250 
then:
L = [10 +/- sqrt(100 - 4*0.1*250)] / 2*0.1 =
  = [10 +/- sqrt(100-100)] / 0.2 =
  = 10/0.2 =
  = 50 hours
The "Labour hours needed to do the job" are 50; then Costs are:
Cost = 50 * $12 = $600

Costs are greater than expected revenues, then CD must reject this offer.

                ----------------------

2. Suppose CD receives a price of $2 per meter, what is CD?s maximizing profit?

To maximize profit, CD firm hire labour until the marginal revenue
product of labour (MRP) equals the wage rate:
MRP = $12

Recall that:
MRP = Marginal Product (MP) * product price ; 

and MP is the derivative of the total product function, then we have:
MP = (10*L - 0.1*L^2)' = 10 - 0.2*L

To maximize profit:
12 = (10 - 0.2*L)*2 = 20 - 0.4*L ==> L = -8/(-0.4) = 20 hours.
Then:
Q(L) = 10*L - 0.1*L^2 = 
     = 10*20 - 0.1*20^2 = 
     = 200 - 0.1*400 =
     = 200 - 40 =
     = 160 meters

The maximum profit is:
Max_Profit = 160 meters * $2/meter - 20 hours * $12/hour =
           = $320 - $240 =
           = $80

For reference regarding the question 2) please visit the following pages:
"THE DEMAND FOR LABOR" from Dr. Bill Wilkes' page at Athens State
University - Athens, Alabama:
http://www.athens.edu/wilkeww/348CHAPT5-REVISED.notes.doc

"Output as a function of a single input: the total product, marginal
product, and average product functions" from Martin J. Osborne's page
at University of Toronto:
http://www.chass.utoronto.ca/~osborne/2x3/tutorial/MP.HTM

"Lecture Notes 23 August 2005 for: Labour Economics III" at School of
Economic & Business Sciences - University of the Witwatersrand,
Johannesburg:
http://www.wits.ac.za/sebs/downloads/2005/demandforlabouriii.doc


Search strategy:
The problems are solved based on my own knowledge of the topic.
For the references I used the following keywords at google.com:
"labour cost" "maximize profit" "marginal cost"
"wage rate" "revenue" labour
"marginal product" derivative


I hope that this helps you. Feel free to request for a clarification
if you need it.

Regards,
livioflores-ga

very brilliant response